Let's solve the problem step by step: --- ## **Given Data** - **Thickness of pan, \( L \) = 0.27 cm = 0.0027 m** - **Diameter of pan, \( D \) = 17 cm = 0.17 m** - **Power consumed = 970 W** - **Fraction transferred to pan = 90%** - **Inner surface temperature, \( T_{inner} \) = 178°C** - **We need to find the heat flux (\( q'' \)) at the bottom of the pan.** --- ## **Step 1: Calculate the Area of the Pan Bottom** Area, \( A \), is given by: \[ A = \frac{\pi}{4} D^2 \] \[ A = \frac{\pi}{4} (0.17)^2 \approx 0.0227 \text{ m}^2 \] --- ## **Step 2: Find the Total Heat Transferred to the Pan** Given only 90% of 970 W is transferred: \[ Q = 0.90 \times 970 = 873 \text{ W} \] --- ## **Step 3: Calculate the Heat Flux at the Bottom** Heat flux (\( q'' \)) is defined as: \[ q'' = \frac{Q}{A} \] Substitute the values: \[ q'' = \frac{873}{0.0227} \approx 38,480 \text{ W/m}^2 \] --- ## **Step 4: Final Answer** \[ \boxed{q'' = 38,480 \ \text{W/m}^2} \] --- ### **Summary Table** | Parameter | Value | Unit | |----------------------|------------|-----------| | Area \(A\) | 0.0227 | m² | | Heat transferred \(Q\) | 873 | W | | Heat flux \(q''\) | 38,480 | W/m² | **The heat flux at the bottom of the pan is \(\boxed{38,480 \ \text{W/m}^2}\).** If you need further steps (like temperature drop across the pan), let me know!